Evaluate (8/27)^3/2

Original post by supreme

Evaluate

(\frac {8}{27})^\frac {3}{2}

The solution involves recognizing that

(\frac {8}{27})^\frac {3}{2} = \frac {8^{{2/2}+1/2}}{27^{{2/2}+1/2}}

Whilst this seems obvious, I just wouldn't have thought of it.

a) Is there any other way of doing it, since an alternative solution given was

\frac {16\sqrt{2/3}}{81}

.

b) How can I show that

\frac {16 \sqrt 6}{243} = \frac {16\sqrt{2/3}}{81}\mathrm{?}

(\dfrac{8}{27})^{\frac{3}{2}}

=(\dfrac{2^3}{3^3})^{\frac{3}{2}}

=(\dfrac{2}{3})^{\frac{9}{2}}

=\dfrac{2^{4+0.5}}{3^{4+0.5}}

=\dfrac{16\sqrt2}{81\sqrt3}

=\dfrac{16\sqrt2/\sqrt3}{81}

=\dfrac{16\sqrt{2/3}}{81}

Reply 2

FireGarden

14

Original post by supreme

Evaluate

(\frac {8}{27})^\frac {3}{2}

The solution involves recognizing that

(\frac {8}{27})^\frac {3}{2} = \frac {8^{{2/2}+1/2}}{27^{{2/2}+1/2}}

Whilst this seems obvious, I just wouldn't have thought of it.

a) Is there any other way of doing it, since an alternative solution given was

\frac {16\sqrt{2/3}}{81}

.

b) How can I show that

\frac {16 \sqrt 6}{243} = \frac {16\sqrt{2/3}}{81}\mathrm{?}

a) Just.. do it.

(\dfrac{8}{27})^{\frac{3}{2}} = (\dfrac{2\sqrt{2}}{3\sqrt{3}})^3 = \dfrac{8\cdot 2\sqrt{2}}{27\cdot 3\sqrt{3}} = \dfrac{16\sqrt{2}}{81\sqrt{3}} = \dfrac{16\sqrt{\frac{2}{3}}}{81}

b) multiply by 3/3.

Reply 3

Original post by supreme

Evaluate

(\frac {8}{27})^\frac {3}{2}

The solution involves recognizing that

(\frac {8}{27})^\frac {3}{2} = \frac {8^{{2/2}+1/2}}{27^{{2/2}+1/2}}

Whilst this seems obvious, I just wouldn't have thought of it.

a) Is there any other way of doing it, since an alternative solution given was

\frac {16\sqrt{2/3}}{81}

.

b) How can I show that

\frac {16 \sqrt 6}{243} = \frac {16\sqrt{2/3}}{81}\mathrm{?}

b)

\dfrac{16\sqrt6}{243}

=\dfrac{16\sqrt6/3}{81}

=\dfrac{16\sqrt{2\times3/9}}{81}

=\dfrac{16\sqrt{2/3}}{81}

Reply 4

OP

Original post by lizard54142

b)

\dfrac{16\sqrt6}{243}

=\dfrac{16\sqrt6/3}{81}

=\dfrac{16\sqrt{2\times3/9}}{81}

=\dfrac{16\sqrt{2/3}}{81}

I am really lost with how you've gone from the top to the second line:-
If we divide 243 by 3 to get 81, why is the same not done to 16? i.e. 16/3?

Reply 5

OP

Thanks, Lizard54142 & FireGarden.

The bit I'm struggling with is how we go from:-

\dfrac{16\sqrt{2}}{81\sqrt{3}} = \dfrac{16\sqrt{\frac{2}{3}}}{81}

Equally, the same problem with:-

\dfrac {16\sqrt6}{243} = \dfrac{16\sqrt{\frac{2}{3}}}{81}

Reply 6

Original post by supreme

I am really lost with how you've gone from the top to the second line:-
If we divide 243 by 3 to get 81, why is the same not done to 16? i.e. 16/3?

I have done that...

\frac{16\sqrt6}{3}

=\frac{16\sqrt6}{\sqrt9}

={16\sqrt{6/9}}

Because

\frac{\sqrt a}{\sqrt b} = \sqrt{a/b}

(edited 8 years ago)

Reply 7

OP

Original post by lizard54142

I have done that...

\frac{16\sqrt6}{3}

=\frac{16\sqrt6}{\sqrt9}

={16\sqrt{6/9}}

Because

\frac{\sqrt a}{\sqrt b} = \sqrt{a/b}

I am sure you did but it's my understanding that's the problem......

So how does

\sqrt 9

come into it? How was that calculated?

Reply 8

Original post by supreme

I am sure you did but it's my understanding that's the problem......

So how does

\sqrt 9

come into it? How was that calculated?

I just substituted it for 3:

3 = \sqrt9

Reply 9

OP

Original post by lizard54142

I just substituted it for 3:

3 = \sqrt9

Just about get it, thanks.

Reply 10

Zacken

22

Original post by supreme

Thanks, Lizard54142 & FireGarden.

The bit I'm struggling with is how we go from:-

\dfrac{16\sqrt{2}}{81\sqrt{3}} = \dfrac{16\sqrt{\frac{2}{3}}}{81}

Equally, the same problem with:-

\dfrac {16\sqrt6}{243} = \dfrac{16\sqrt{\frac{2}{3}}}{81}

The first one is easy

\displaystyle \frac{16\sqrt{2}}{81\sqrt{3}} = \frac{16}{81} \times \frac{\sqrt{2}}{\sqrt{3}}

.

Using

\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}

, we get

\displaystyle \frac{16}{81} \times \sqrt{\frac{2}{3}}

.

But you also know that

\displaystyle \frac{16}{81} \times a = \frac{16a}{81}

.

So you get

\displaystyle \frac{16}{81} \times \sqrt{\frac{2}{3}} = \frac{16\sqrt{\frac{2}{3}}}{81}

.

For the next part,

\displaystyle \frac{16\sqrt{6}}{243} = \frac{16\sqrt{6}}{81} \times \frac{1}{3}=\frac{16}{81} \times \frac{\sqrt{6}}{3}

.

But you know that you can re-write

3 = \sqrt{9}

so it becomes

\displaystyle \frac{16}{81} \times \frac{\sqrt{6}}{\sqrt{9}} = \frac{16}{81} \times \sqrt{\frac{6}{9}}

But

\displaystyle \frac{6}{9} = \frac{2}{3}

So it becomes

\displaystyle \frac{16}{81} \times \sqrt{\frac{2}{3}} = \frac{16\sqrt{\frac{2}{3}}}{81}

.

Geddit?

Reply 11

OP